Problem 58 - Thomas Calculus

Question

Find the limits in Exercises \(51-66\) $$ \lim _{x \rightarrow 0}\left(e^{x}+x\right)^{1 / x} $$

Step-by-Step Solution

Answer

The limit is \(e^2\).

Step 1: Understand the Expression

The expression given is \( \lim_{x \rightarrow 0}(e^{x} + x)^{1/x} \). We need to find the limit of this expression as \(x\) approaches 0.

Step 2: Apply Logarithm

To simplify the limit, take the natural logarithm of the expression: \( \ln{y} = \lim_{x \rightarrow 0}\frac{\ln(e^{x} + x)}{x} \), where \(y = (e^{x} + x)^{1/x} \).

Step 3: Use L'Hospital's Rule

The expression for \(\ln{y}\) turns into an indeterminate form \(\frac{0}{0}\) as \(x \rightarrow 0\). Apply L'Hospital's Rule: differentiate the numerator and denominator separately.

Step 4: Differentiate Numerator and Denominator

Differentiate \(\ln(e^{x} + x)\) with respect to \(x\), which gives \(\frac{e^{x}+1}{e^{x}+x}\), and differentiate \(x\), which gives 1.

Step 5: Simplify Limit

Now compute the limit: \( \lim_{x \rightarrow 0} \frac{e^{x} + 1}{e^{x} + x} \). When \(x\rightarrow 0\), the expression simplifies to \(\frac{2}{1}\) or just 2.

Step 6: Exponentiate to Find Original Limit

Since we found \(\ln{y}\), we have \(\ln{y} = 2\) implying \(y = e^{2}\). Therefore, the original limit is \(\lim_{x \rightarrow 0}(e^{x} + x)^{1/x} = e^{2}\).

Key Concepts

L'Hospital's Rule

L'Hospital's Rule is a powerful method for finding limits of indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). This technique assists in determining the limit of a fraction by differentiating the numerator and denominator separately, potentially simplifying the expression.
For instance, consider the limit \(\lim_{x \rightarrow a} \frac{f(x)}{g(x)}\). If this results in an indeterminate form, apply L'Hospital's Rule: differentiate \(f(x)\) to get \(f'(x)\) and \(g(x)\) to get \(g'(x)\). Then re-evaluate the limit \(\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}\).

• Key requirements are that \(f(x)\) and \(g(x)\) need to be differentiable around the neighborhood of \(a\).
• The derivative \(g'(x)\) must not be zero at \(a\).

Using L'Hospital's Rule can simplify complex limits, such as the one in our exercise, by transforming difficult expressions into manageable forms.

Exponential Functions

Exponential functions are fundamental in calculus and beyond, typically expressed with the base \(e\) (Euler's number, approximately 2.718). These functions are of the form \(e^{x}\), where \(x\) is any real number.
What makes exponential functions so special? They have distinctive properties such as rapid growth and a unique rate of change - the function's rate of growth is proportional to its current value.

• This characteristic means that the derivative of \(e^{x}\) is also \(e^{x}\).
• This quality makes them indispensable in problems involving growth and decay, like population studies and radioactive decay.

Exponential functions can be manipulated using logarithms, allowing expressions to be handled in different mathematical contexts. For example, in our exercise, the expression \(e^{x}\) coincides with the natural logarithm \(\ln\) for simplification purposes.

Natural Logarithms

The natural logarithm is a logarithm with base \(e\), written as \(\ln(x)\). It is the inverse operation of raising \(e\) to a power. The natural logarithm is particularly useful in calculus because of its set of unique properties, including:

• The derivative of \(\ln(x)\) is \(1/x\), a fact often utilized in deriving solutions in calculus.
• It simplifies exponential expressions, making them easier to manipulate.

In our exercise, taking \(\ln\) of the original expression \((e^{x}+x)^{1/x}\) simplifies handling the power of \(1/x\). This transformation gives form to the problem making it suitable for applications like L'Hospital's Rule.
Understanding the natural logarithm paves the way to comprehending growth behavior, exponential decay, and more sophisticated calculus operations.

Indeterminate Forms

Indeterminate forms are specific limit expressions that do not initially reveal their value, like \(\frac{0}{0}\), \(\infty/\infty\), or \(0 \cdot \infty\). These expressions hint that the straightforward application of limit properties won't work directly.
These forms demand special techniques like L'Hospital's Rule or algebraic manipulation to evaluate. They occur when substituting a value into a function results in ambiguous forms.

• Common indeterminate forms include \(\frac{0}{0}\) and \(\infty - \infty\).
• Careful analysis often requires rewriting the expression, sometimes using derivatives or factorizations.

In our exercise, as \(x\) approaches 0, both the numerator and the denominator go to 0, resulting in an indeterminate \(\frac{0}{0}\) form. This scenario highlights the necessity for approaches like L'Hospital's Rule, enabling further simplification and solution.